A hot body,obeying Newton's law of cooling,is | vedclass

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(B) According to Newton's law of cooling,the time $t$ taken for a body to cool from temperature $	heta_1$ to $	heta_2$ in an environment with ambien

Vedclass · Accepted Answer

$8.6$

Vedclass · Answer

(B) According to Newton's law of cooling,the time $t$ taken for a body to cool from temperature $	heta_1$ to $	heta_2$ in an environment with ambient temperature $	heta_0$ is given by $t = \frac{1}{k} \ln \left( \frac{	heta_1 - 	heta_0}{	heta_2 - 	heta_0} ight)$.For the first interval ($80\,^oC$ to $40\,^oC$): $5 = \frac{1}{k} \ln \left( \frac{80 - 30}{40 - 30} ight) = \frac{1}{k} \ln \left( \frac{50}{10} ight) = \frac{1}{k} \ln 5$. So,$k = \frac{\ln 5}{5} = \frac{1.609}{5} = 0.3218\,min^{-1}$.For the second interval ($62\,^oC$ to $32\,^oC$): $t = \frac{1}{k} \ln \left( \frac{62 - 30}{32 - 30} ight) = \frac{1}{k} \ln \left( \frac{32}{2} ight) = \frac{1}{k} \ln 16 = \frac{1}{k} \ln(2^4) = \frac{4 \ln 2}{k}$.Substituting the values: $t = \frac{4 	imes 0.693}{0.3218} \approx \frac{2.772}{0.3218} \approx 8.61\,minutes$.Thus,the time taken is $8.6\,minutes$.

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